Multiply the following complex numbers: $({-3-3i}) \cdot ({-2-i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-3-3i}) \cdot ({-2-i}) = $ $ ({-3} \cdot {-2}) + ({-3} \cdot {-1}i) + ({-3}i \cdot {-2}) + ({-3}i \cdot {-1}i) $ Then simplify the terms: $ (6) + (3i) + (6i) + (3 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 6 + (3 + 6)i + 3i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 6 + (3 + 6)i - 3 $ The result is simplified: $ (6 - 3) + (9i) = 3+9i $